Question

$\int \frac{dx}{\sin x - \cos x + \sqrt{2}} $ equals to

• A. $ - \frac{1}{\sqrt{2}} \tan \left(\frac{x}{2} + \frac{\pi}{8}\right) + C $
• B. $ \frac{1}{2} \tan \left(\frac{x}{2} + \frac{\pi}{8}\right) + C $
• C. $ \frac{1}{\sqrt{2}} \cot\left(\frac{x}{2} + \frac{\pi}{8}\right) + C $
• D. $ - \frac{1}{\sqrt{2}} \cot \left(\frac{x}{2} + \frac{\pi}{8}\right) + C $

Answer 1

Answer: (D)

Let $I = \int \frac{dx}{\sin x - \cos x + \sqrt{2}} $
$= \int \frac{dx}{\sqrt{2}\left(\sin x \sin \frac{\pi}{4} - \cos x \cos \frac{\pi}{4} + 1\right)} $
$= \frac{1}{\sqrt{2}}\int \frac{dx}{1-\cos\left(x + \frac{\pi}{4}\right)} $
$= \frac{1}{\sqrt{2}}\int \frac{dx}{2 \sin^{2} \left(\frac{x}{2} + \frac{\pi }{8}\right)} $
$= \frac{1}{2\sqrt{2}}\int cosec^{2}\left(\frac{x}{2} + \frac{\pi }{8}\right)dx$
$ = \frac{1}{2\sqrt{2}} \frac{-\cot \left(\frac{x}{2}+ \frac{\pi }{8}\right)}{\frac{1}{2}} +C$
$ = -\frac{1}{\sqrt{2}}\cot\left(\frac{x}{2}+\frac{\pi}{8}\right)+C $